已知数列 $\left\{ {{a_n}} \right\}$ 满足 ${a_0}=\dfrac{1}{2}$，${a_n}={a_{n-1}}+\dfrac{1}{{{n^2}}} \cdot a_{n-1}^2$，求证：$\dfrac{{n+1}}{{n+2}}<{a_n}<n$．
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【答案】
略
【解析】
注意到 $\left\{ {\dfrac{1}{{{a_n}}}} \right\}$ 单调递减趋于 $0$，将递推式改写为$$\dfrac{1}{{{a_{n-1}}}}-\dfrac{1}{{{a_n}}}=\dfrac{1}{{{a_{n-1}}}}-\dfrac{1}{{{a_{n-1}}+\dfrac{1}{{{n^2}}} \cdot {a_{n-1}}^2}}=\dfrac{1}{{{n^2}+{a_{n-1}}}}$$欲证不等式为$$\dfrac{1}{n}<\dfrac{1}{{{a_n}}}<\dfrac{{n+2}}{{n+1}},$$也即$$\dfrac{n}{{n+1}}<\dfrac{1}{{{a_0}}}-\dfrac{1}{{{a_n}}}<2-\dfrac{1}{n}.$$由 ${a_{n-1}}>0$ 得$$\dfrac{1}{{{a_{n-1}}}}-\dfrac{1}{{{a_n}}}<\dfrac{1}{{{n^2}}}$$于是$$\dfrac{1}{{{a_0}}}-\dfrac{1}{{{a_n}}}=\sum\limits_{k=1}^n {\left( {\dfrac{1}{{{a_{k-1}}}}-\dfrac{1}{{{a_k}}}} \right)}<\sum\limits_{k=1}^n {\dfrac{1}{{{k^2}}}}<1+\sum\limits_{k=2}^n {\dfrac{1}{{\left( {k-1} \right)k}}}=2-\dfrac{1}{n}$$此时 ${a_n}<n$，于是$$\dfrac{1}{{{a_{n-1}}}}-\dfrac{1}{{{a_n}}}=\dfrac{1}{{{n^2}+{a_{n-1}}}}>\dfrac{1}{{{n^2}+n-1}}>\dfrac{1}{n}-\dfrac{1}{{n+1}}$$进而$$\dfrac{1}{{{a_0}}}-\dfrac{1}{{{a_n}}}=\sum\limits_{k=1}^n {\left( {\dfrac{1}{{{a_{k-1}}}}-\dfrac{1}{{{a_k}}}} \right)}>\sum\limits_{k=1}^n {\left( {\dfrac{1}{k}-\dfrac{1}{{k+1}}} \right)}=\dfrac{n}{{n+1}}.$$因此原不等式成立．
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